Explanation:

We can solve this problem by using the equation of motion:

s = ut + 1/2 at^2

where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Let's assume that the first body is dropped from rest, so its initial velocity is zero. We can use the same equation for both bodies, but we need to find the values of u and a separately.

Body 1:
- u1 = 0 (initial velocity)
- a1 = 9.8 m/s^2 (acceleration due to gravity)

Body 2:
- u2 = 2 m/s (initial velocity)
- a2 = 9.8 m/s^2 (acceleration due to gravity)

We know that the separation between the bodies after time t is 18 m. This means that the distance traveled by each body is half of that, or 9 m.

Now we can use the equation of motion for each body to find the time taken to cover 9 m.

For body 1:
9 = 0*t + 1/2*9.8*t^2
t^2 = 18/9.8
t = sqrt(1.8367)
t ≈ 1.354 s

For body 2:
9 = 2*t + 1/2*9.8*t^2
9.8t^2 + 2t - 18 = 0
Using the quadratic formula, we get:
t = (-2 ± sqrt(2^2 - 4*9.8*(-18))) / (2*9.8)
t ≈ 7.645 s or t ≈ -0.945 s
Since we are looking for the positive time, we take t ≈ 7.645 s.

Finally, we can see that the time taken for both bodies to cover the separation distance of 18 m is the same, which is t = 7.645 s. However, this is the time taken by the second body, which was thrown downward. The first body, which was dropped, takes an additional time of 1.354 s to cover the same distance. Therefore, the total time taken is:

t_total = 7.645 s + 1.354 s = 9 s

Answer: The total time taken for both bodies to cover the separation distance of 18 m is 9 s.